标签：subarray which nbsp ems [] not bsp btn tput
Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.
Example 1:
Input:nums = [1,1,1], k = 2 Output: 2
Note:
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笨方法就是遍历所有子序列，由于起点和终点都要遍历，所以时间复杂度是O（n^2）
本次的方法是遍历所有起点是最开始元素的子序列，我们起名叫sum[i]，只要两个sum[i]的差值是k，则两个i作为起点和终点的子序列就符合要求。由于起点就是最开始元素，只需遍历终点，时间复杂度为O（n）。
public class Solution {
public int subarraySum(int[] nums, int k) {
Map<Integer,Integer> map = new HashMap<Integer,Integer>();
map.put(0,1);
int sum = 0;
int res = 0;
for(int i = 0;i<nums.length;i++)
{
sum += nums[i];
res += map.getOrDefault(sum-k,0);
map.put(sum,map.getOrDefault(sum,0)+1);
}
return res;
}
}
[leetcode 560. Subarray Sum Equals K]
标签：subarray which nbsp ems [] not bsp btn tput
原文地址：http://www.cnblogs.com/stAr-1/p/6817838.html
